"Diez B. Roggisch" <[EMAIL PROTECTED]> writes:
> bruce schrieb:
> > hi...
> > i can do the following
> > br = Browser
> > br.open("www.yahoo.com")
> > br.open("file:///foo")
> > but can i do
> > s = "..." qualified html text
> > br.open(s)
> > i'm curious, if i have html from some
bruce schrieb:
> hi...
>
> i can do the following
> br = Browser
> br.open("www.yahoo.com")
> br.open("file:///foo")
>
> but can i do
> s = "..." qualified html text
> br.open(s)
>
> i'm curious, if i have html from someother source, is there a way to simply
> get it into the "B
hi...
i can do the following
br = Browser
br.open("www.yahoo.com")
br.open("file:///foo")
but can i do
s = "..." qualified html text
br.open(s)
i'm curious, if i have html from someother source, is there a way to simply
get it into the "Browser" so i can modify it...
thanks
-b
bruce wrote:
> r2 = br.follow_link(url_regex=re.compile(r"\*"),nr=1) <<<
Seems to me your regex is buggy. You are using a "raw string" yet you
still escape the asterisk ('*') with a backslash? This will only match
a string which contains an asterisk, while I'm guessing you're trin
hi..
i have the following piece of test code. i'm trying to implement/check out
the follow-link method. i'm just trying to figure out how to get a link from
the page.
i was hoping that the regex would basically get the 1st url link...
any thoughts/comments/ideas as to what i'm doing wrong.
than
